Investigate if y=x^99 + 99^x is increasin or decreasing.

In order to prove that f(x) is an increasing function,
we'll do the first derivative test.

If the first derivative
of the function is positive, then the function is
increasing.

If the first derivative of the function is
negative, then the function is decreasing.

Let's calculate
f'(x):

 f'(x) = (x^99 +
99^x)'

 f'(x) = (x^99)' +
(99^x)'

 f'(x) = 99x^98 +
99^x*ln99

Since x^98 is positive for any real value of x,
then 99x^98 is also positive.

We'll calculate ln99 = 4.595
approx. (1)

99^x >0
(2)

From (1) and (2) =>
99^x*ln99>0

f'(x)=99x^98 +
99^x*ln99>0

Since each
term of the expression of f'(x) is positive, the sum of positive terms is also a
positive expression. The expression of f'(x) it's obviously>0, so f(x) is an
increasing function.